# Dispersion of light

Dispersion:

The dispersion is the splitting up white light in to its constituent colors.

Cause of dispersion;

The cause of dispersion of light is due to the variation of the refractive index of the medium with color of light.

Dispersive power related to refractive index of the material;

Dispersive power can be defined as the relative change in the focal length of a lens$\frac{{{\rm{df}}}}{{\rm{f}}}$. Here df is the change in focal length and dμ is the corresponding change in the refractive index of the material of the lens.  The dispersive power for a small angle prism is defined as the ratio of the angular dispersion to the deviation produced by it to mean light. I.e. ${\rm{\: }}\omega = {\rm{\: }}\frac{{{\mu _{\rm{v}}} - {\mu _{\rm{r}}}}}{{\mu - 1}}$

A mirror cannot give rise to chromatic aberration;

When a ray of light is incident on a mirror, reflection takes place but not refraction. As the reflection of light is in same medium, the velocity of different colors of light is same. This implies that a mirror cannot give rise to chromatic aberration.

Chromatic aberration;

The inability of lens to focus all colors of light at a single point is called the chromatic aberration. It is removed by combining two lenses.

Blue light is deviated more than red light by prism:

The deviation produced by a small angle prism is $\delta = {\rm{A}}\left( {\mu - 1} \right)$ where A is the angle of prism and μ is the refractive index of prism material. The refractive index increases with decrease of wavelength. Since the wavelength of blue light is less than that of red light, the blue light has more refractive index than red light. Hence, blue light is deviated more than red light by prism.

Dispersion is caused by refraction and not by reflection;

The dispersion occurs due to refraction because the white light has to travel in the glass medium to disperse whereas the mirror is an opaque object.

Dispersive medium:

A substance in which waves of different frequencies travel at different speed is known as dispersive medium

Relation for the dispersive power;

The power of a transparent medium to separate different colors of light by refraction as measured by the difference in refractivity for two specified widely differing wavelengths divided by the refractivity at some specified intermediate wavelength.

Dispersive power, ω= $\frac{{{\delta _{\rm{v}}} - {\delta _{\rm{r}}}}}{\delta }$

$\omega = {\rm{\: }}\frac{{{\rm{A}}\left( {{\mu _{\rm{v}}} - 1} \right) - {\rm{A}}\left( {{\mu _{\rm{r}}} - 1} \right)}}{{{\rm{A}}\left( {\mu - 1} \right)}}$

Here, δ =A (μ -1) is angle of deviation for mean color.

$\omega = {\rm{\: }}\frac{{{\rm{A}}{\mu _{\rm{v}}} - {\rm{A}} - {\rm{A}}{\mu _{\rm{r}}} + {\rm{A}}}}{{{\rm{A}}\left( {\mu - 1} \right)}}$

Or, ${\rm{\: }}\omega = {\rm{\: }}\frac{{{\rm{A}}\left( {{\mu _{\rm{v}}} - {\mu _{\rm{v}}}} \right)}}{{{\rm{A}}\left( {\mu - 1} \right)}}$

Or, ${\rm{\: }}\omega = {\rm{\: }}\frac{{{\mu _{\rm{v}}} - {\mu _{\rm{r}}}}}{{\mu - 1}}$

This the relation for the dispersive power

Angular dispersion and dispersive power;

Angular dispersion is a measure of the angular separation of light rays of different wavelength or color traversing a prism or diffraction grating, equal to the rate of change of the angle of deviation with respect to the change in wavelength.

Dispersive power is the power of a transparent medium to separate different colors of light by refraction as measured by the difference in refractivity for two specified widely differing wavelengths divided by the refractivity at some specified intermediate wavelength.

Dispersive power can be defined as the relative change in the focal length of a lens$\frac{{{\rm{df}}}}{{\rm{f}}}$. Here df is the change in focal length and dμ is the corresponding change in the refractive index of the material of the lens.  The dispersive power for a small angle prism is defined as the ratio of the angular dispersion to the deviation produced by it to mean light. I.e.

${\rm{\: }}\omega = {\rm{\: }}\frac{{{\mu _{\rm{v}}} - {\mu _{\rm{r}}}}}{{\mu - 1}}$

The angular dispersion between two colors of light is defined as the difference in deviation produced by a prism between the two colors of light.

The condition to produce deviation without dispersion in a combination of prisms; The emergent ray is undispersed. For this

$\frac{{{\rm{A'}}}}{{\rm{A}}} =$ $\frac{{{\mu _{\rm{v}}} - {\mu _{\rm{R}}}}}{{\mu _{\rm{v}}^{\rm{'}} - {\mu _{\rm{R}}}}}$
This is the condition for achromatism or no dispersion.

Let µ'V, µ'R and µ be the corresponding values for the material of the second prism, while µV, µR & µ be of the first prism.

Angular deviation by first prism is
V - δ R) = (µV - µR)A

Angular dispersion produced by second prism
(δ 'V - δ 'R) = (µ'V - µ'R)A'

Since combination does not produce any dispersion
(δ V - δ R) + (δ 'V - δ 'R) = 0

or A(µV - µR) + A'(µ'V - µ'R) = 0

Or  $\frac{{{\rm{A'}}}}{{\rm{A}}} =$ $\frac{{{\mu _{\rm{v}}} - {\mu _{\rm{R}}}}}{{\mu _{\rm{v}}^{\rm{'}} - {\mu _{\rm{R}}}}}$

Dispersion without deviation: Let two prisms of angle A & A' be placed as shown in figure. If the system produces net deviation as zero
then

$\frac{{{\rm{A'}}}}{{\rm{A}}} =$ $\frac{{\left( {\mu - 1} \right)}}{{\left( {\mu - 1} \right)}}$

Where µ & µ' are the refractive index of mean wavelength.

Deviation without dispersion; The emergent ray is undispersed

For this

$\frac{{{\rm{A'}}}}{{\rm{A}}} =$ $\frac{{{\mu _{\rm{v}}} - {\mu _{\rm{R}}}}}{{\mu _{\rm{v}}^{\rm{'}} - {\mu _{\rm{R}}}}}$
This is the condition for achromatism or no dispersion.

Let µ'V, µ'R and µ be the corresponding values for the material of the second prism, while µV, µR & µ be of the first prism.

Angular deviation by first prism is
V - δ R) = (µV - µR)A

Angular dispersion produced by second prism
(δ 'V - δ 'R) = (µ'V - µ'R)A'

Since combination does not produce any dispersion
(δ V - δ R) + (δ 'V - δ 'R) = 0

or A(µV - µR) + A'(µ'V - µ'R) = 0

Or  $\frac{{{\rm{A'}}}}{{\rm{A}}} =$ $\frac{{{\mu _{\rm{v}}} - {\mu _{\rm{R}}}}}{{\mu _{\rm{v}}^{\rm{'}} - {\mu _{\rm{R}}}}}$

Chromatic aberration and expression for longitudinal aberration in case of thin lens:

When a ray of light is incident on a mirror, reflection takes place but not refraction. As the reflection of light is in same medium, the velocity of different colors of light is same. This implies that a mirror cannot give rise to chromatic aberration.

If ${{\rm{f}}_{\rm{r}}}$ and ${{\rm{f}}_{\rm{V}}}$ be the focal length of the lens for the red and violet light, longitudinal chromatic aberration = ${{\rm{f}}_{\rm{r}}} - {{\rm{f}}_{\rm{v}}}.$

If ${{\rm{R}}_1}{\rm{and\: }}{{\rm{R}}_2}{\rm{be\: }}$ the radii of curvature of the surfaces of the lens, and μ is refractive index

$\frac{1}{{\rm{f}}} = \left( {\mu - 1} \right)\left( {\frac{1}{{{{\rm{R}}_1}}} + \frac{1}{{{{\rm{R}}_2}}}} \right)$

$\frac{1}{{{{\rm{R}}_1}}} + \frac{1}{{{{\rm{R}}_2}}}{\rm{\: }} = \frac{1}{{{\rm{f}}\left( {\mu - 1} \right)}} \ldots \ldots \ldots \ldots \ldots ..1$

${\mu _{\rm{r}}} = {\rm{\: }}$Refractive index of the lens for red light

${\mu _{\rm{v}}} = {\rm{\: \: }}$Refractive index of the lens for violet light

$\frac{1}{{{{\rm{f}}_{\rm{r}}}}} = \left( {{\mu _{\rm{r}}} - 1} \right)\left( {\frac{1}{{{{\rm{R}}_1}}} + \frac{1}{{{{\rm{R}}_2}}}} \right)$………………………..2

$\frac{1}{{{{\rm{f}}_{\rm{v}}}}} = \left( {{\mu _{\rm{v}}} - 1} \right)\left( {\frac{1}{{{{\rm{R}}_1}}} + \frac{1}{{{{\rm{R}}_2}}}} \right)$………………………..3

∴$\frac{1}{{{{\rm{f}}_{\rm{v}}}}} - \frac{1}{{{{\rm{f}}_{\rm{r}}}}} = \left( {\frac{1}{{{{\rm{R}}_1}}} + \frac{1}{{{{\rm{R}}_2}}}} \right)\left( {{\mu _{\rm{v}}} - 1 - {\mu _{\rm{r}}} + 1} \right)$

$= \left( {\frac{1}{{{{\rm{R}}_1}}} + \frac{1}{{{{\rm{R}}_2}}}} \right)\left( {{\mu _{\rm{v}}} - {\mu _{\rm{r}}}} \right)$

∴$\frac{1}{{{{\rm{f}}_{\rm{v}}}}} - \frac{1}{{{{\rm{f}}_{\rm{r}}}}} = \frac{{{\mu _{\rm{v}}} - {\mu _{\rm{r}}}}}{{{\rm{f}}\left( {{\mu _{\rm{r}}} - 1} \right)}}$

Using 1

$\frac{{{{\rm{f}}_{\rm{r}}} - {{\rm{f}}_{\rm{v}}}}}{{{{\rm{f}}_{{\rm{r}}.{{\rm{f}}_{\rm{v}}}}}}} = \frac{{{\mu _{\rm{v}}} - {\mu _{\rm{r}}}}}{{{\rm{f}}\left( {{\mu _{\rm{r}}} - 1} \right)}}$

${{\rm{f}}_{\rm{r}}} - {{\rm{f}}_{\rm{v}}} = \frac{{\rm{w}}}{{\rm{f}}}{\rm{*}}{{\rm{f}}_{\rm{r}}}.{{\rm{f}}_{\rm{v}}}$

$\frac{{{\mu _{\rm{v}}} - {\mu _{\rm{r}}}}}{{\left( {{\mu _{\rm{r}}} - 1} \right)}}$ = w is the dispersive power of the lens.

If f is the focal length of the lens for mean light, then, we can take

${\rm{f}} = \sqrt {{{\rm{f}}_{\rm{v}}}{{\rm{f}}_{\rm{r}}}}$

${{\rm{f}}_{\rm{r}}} - {{\rm{f}}_{\rm{v}}} = \frac{{\rm{w}}}{{\rm{f}}}{\rm{*}}{{\rm{f}}^2} = {\rm{wf}}$

Hence the longitudinal chromatic aberration of a lens is given by
${{\rm{f}}_{\rm{r}}} - {{\rm{f}}_{\rm{v}}} = {\rm{wf}}$= dispersive power * focal length for mean light.

For a lens the chromatic aberration is the product of dispersive power and its mean focal length;

When a ray of light is incident on a mirror, reflection takes place but not refraction. As the reflection of light is in same medium, the velocity of different colors of light is same. This implies that a mirror cannot give rise to chromatic aberration.

If ${{\rm{f}}_{\rm{r}}}$ and ${{\rm{f}}_{\rm{V}}}$ be the focal length of the lens for the red and violet light, longitudinal chromatic aberration = ${{\rm{f}}_{\rm{r}}} - {{\rm{f}}_{\rm{v}}}.$

If ${{\rm{R}}_1}{\rm{and\: }}{{\rm{R}}_2}{\rm{be\: }}$ the radii of curvature of the surfaces of the lens, and μ is refractive index

$\frac{1}{{\rm{f}}} = \left( {\mu - 1} \right)\left( {\frac{1}{{{{\rm{R}}_1}}} + \frac{1}{{{{\rm{R}}_2}}}} \right)$

$\frac{1}{{{{\rm{R}}_1}}} + \frac{1}{{{{\rm{R}}_2}}}{\rm{\: }} = \frac{1}{{{\rm{f}}\left( {\mu - 1} \right)}} \ldots \ldots \ldots \ldots \ldots ..1$

${\mu _{\rm{r}}} = {\rm{\: }}$Refractive index of the lens for red light

${\mu _{\rm{v}}} = {\rm{\: \: }}$Refractive index of the lens for violet light

$\frac{1}{{{{\rm{f}}_{\rm{r}}}}} = \left( {{\mu _{\rm{r}}} - 1} \right)\left( {\frac{1}{{{{\rm{R}}_1}}} + \frac{1}{{{{\rm{R}}_2}}}} \right)$………………………..2

$\frac{1}{{{{\rm{f}}_{\rm{v}}}}} = \left( {{\mu _{\rm{v}}} - 1} \right)\left( {\frac{1}{{{{\rm{R}}_1}}} + \frac{1}{{{{\rm{R}}_2}}}} \right)$………………………..3

∴$\frac{1}{{{{\rm{f}}_{\rm{v}}}}} - \frac{1}{{{{\rm{f}}_{\rm{r}}}}} = \left( {\frac{1}{{{{\rm{R}}_1}}} + \frac{1}{{{{\rm{R}}_2}}}} \right)\left( {{\mu _{\rm{v}}} - 1 - {\mu _{\rm{r}}} + 1} \right)$

$= \left( {\frac{1}{{{{\rm{R}}_1}}} + \frac{1}{{{{\rm{R}}_2}}}} \right)\left( {{\mu _{\rm{v}}} - {\mu _{\rm{r}}}} \right)$

∴$\frac{1}{{{{\rm{f}}_{\rm{v}}}}} - \frac{1}{{{{\rm{f}}_{\rm{r}}}}} = \frac{{{\mu _{\rm{v}}} - {\mu _{\rm{r}}}}}{{{\rm{f}}\left( {{\mu _{\rm{r}}} - 1} \right)}}$

Using 1

$\frac{{{{\rm{f}}_{\rm{r}}} - {{\rm{f}}_{\rm{v}}}}}{{{{\rm{f}}_{{\rm{r}}.{{\rm{f}}_{\rm{v}}}}}}} = \frac{{{\mu _{\rm{v}}} - {\mu _{\rm{r}}}}}{{{\rm{f}}\left( {{\mu _{\rm{r}}} - 1} \right)}}$

${{\rm{f}}_{\rm{r}}} - {{\rm{f}}_{\rm{v}}} = \frac{{\rm{w}}}{{\rm{f}}}{\rm{*}}{{\rm{f}}_{\rm{r}}}.{{\rm{f}}_{\rm{v}}}$

$\frac{{{\mu _{\rm{v}}} - {\mu _{\rm{r}}}}}{{\left( {{\mu _{\rm{r}}} - 1} \right)}}$ = w is the dispersive power of the lens.

If f is the focal length of the lens for mean light, then, we can take

${\rm{f}} = \sqrt {{{\rm{f}}_{\rm{v}}}{{\rm{f}}_{\rm{r}}}}$

${{\rm{f}}_{\rm{r}}} - {{\rm{f}}_{\rm{v}}} = \frac{{\rm{w}}}{{\rm{f}}}{\rm{*}}{{\rm{f}}^2} = {\rm{wf}}$

Hence the longitudinal chromatic aberration of a lens is given by
${{\rm{f}}_{\rm{r}}} - {{\rm{f}}_{\rm{v}}} = {\rm{wf}}$= dispersive power * focal length for mean light.

Achromatism in lenses:

An achromatic lens or achromatic is a lens that is designed to limit the effects of chromatic and spherical aberration. Achromatic lenses are corrected to bring two wavelengths (typically red and blue) into focus in the same plane.

The most common type of achromatic is the achromatic doublet, which is composed of two individual lenses made from glasses with different amounts of dispersion, one element is a negative (concave) element made out of flint, which has relatively high dispersion, and the other is a positive (convex) element made of crown glass, which has lower dispersion. The lens elements are mounted next to each other, often cemented together, and shaped so that the chromatic aberration of one is counter balanced by that of the other.

This is the condition for achromatic combination of two lenses.
$\frac{{\rm{w}}}{{\rm{f}}} + \frac{{\rm{w}}}{{\rm{f}}}$ = 0

The relation can be obtained by the method of calculus.

$\frac{1}{{\rm{f}}} = \left( {\mu - 1} \right)\left( {\frac{1}{{{{\rm{R}}_1}}} + \frac{1}{{{{\rm{R}}_2}}}} \right)$

$\frac{1}{{{{\rm{R}}_1}}} + \frac{1}{{{{\rm{R}}_2}}} = \frac{1}{{\left( {\mu - 1} \right){\rm{f}}}} = {\rm{K}}$

Or $\frac{1}{{\rm{f}}} = {\rm{K}}\left( {\mu - 1} \right)$

Differentiating with respect to f and μ,

$- \frac{{{\rm{dF}}}}{{{{\rm{f}}^2}}} = {\rm{kdf}}\mu = \frac{{{\rm{d}}\mu }}{{\left( {\mu - 1} \right){\rm{f}}}} = \frac{\omega }{{\rm{f}}}$

$\omega = - \frac{{{\rm{df}}}}{{\rm{F}}}$……………..1

Where $\omega$ is the dispersive power of the material of the lens.

From equation 1

Dispersive power can be defined as the relative change in the focal length of a lens$\frac{{{\rm{df}}}}{{\rm{f}}}$. Here df is the change in focal length and dμ is the corresponding change in the refractive index of the material of the lens.

Then,

Focal length F of their combination is given by $\frac{1}{{\rm{F}}} = \frac{1}{{\rm{f}}} + \frac{1}{{\rm{f}}}$

Differentiating, we get

$- \frac{{\delta {\rm{F}}}}{{{{\rm{F}}^2}}} = - \frac{{\delta {\rm{f}}}}{{{{\rm{f}}^2}}} - \frac{{\delta {\rm{f'}}}}{{{{\rm{f}}^{{\rm{'}}2}}}}$

Or, ${\rm{\: }} - \frac{{\delta {\rm{F}}}}{{{{\rm{F}}^2}}} = \frac{\omega }{{\rm{f}}} + \frac{{\omega '}}{{{\rm{f'}}}} \ldots \ldots \ldots \ldots \ldots 2$

Hence, for achromatism i.e. δ f = 0, then

And condition for achromatism is

$\frac{{\rm{w}}}{{\rm{f}}} + \frac{{{\rm{w'}}}}{{\rm{f}}}$ = 0

This is required expression

The relation for achromatism of two thin lenses in contact:

The relation $\frac{{\rm{w}}}{{\rm{f}}} + \frac{{\rm{w}}}{{\rm{f}}}$ = 0 can be obtained by the method of calculus.

$\frac{1}{{\rm{f}}} = \left( {\mu - 1} \right)\left( {\frac{1}{{{{\rm{R}}_1}}} + \frac{1}{{{{\rm{R}}_2}}}} \right)$

$\frac{1}{{{{\rm{R}}_1}}} + \frac{1}{{{{\rm{R}}_2}}} = \frac{1}{{\left( {\mu - 1} \right){\rm{f}}}} = {\rm{K}}$

Or $\frac{1}{{\rm{f}}} = {\rm{K}}\left( {\mu - 1} \right)$

Differentiating with respect to f and μ,

$- \frac{{{\rm{dF}}}}{{{{\rm{f}}^2}}} = {\rm{kdf}}\mu = \frac{{{\rm{d}}\mu }}{{\left( {\mu - 1} \right){\rm{f}}}} = \frac{\omega }{{\rm{f}}}$

$\omega = - \frac{{{\rm{df}}}}{{\rm{F}}}$……………..1

Where $\omega$ is the dispersive power of the material of the lens.

From equation 1

Dispersive power can be defined as the relative change in the focal length of a lens$\frac{{{\rm{df}}}}{{\rm{f}}}$. Here df is the change in focal length and dμ is the corresponding change in the refractive index of the material of the lens.

Then,

Focal length F of their combination is given by $\frac{1}{{\rm{F}}} = \frac{1}{{\rm{f}}} + \frac{1}{{\rm{f}}}$

Differentiating, we get

$- \frac{{\delta {\rm{F}}}}{{{{\rm{F}}^2}}} = - \frac{{\delta {\rm{f}}}}{{{{\rm{f}}^2}}} - \frac{{\delta {\rm{f'}}}}{{{{\rm{f}}^{{\rm{'}}2}}}}$

Or, ${\rm{\: }} - \frac{{\delta {\rm{F}}}}{{{{\rm{F}}^2}}} = \frac{\omega }{{\rm{f}}} + \frac{{\omega '}}{{{\rm{f'}}}} \ldots \ldots \ldots \ldots \ldots 2$

Hence, for achromatism i.e. δ f = 0, then

And condition for achromatism is

$\frac{{\rm{w}}}{{\rm{f}}} + \frac{{{\rm{w'}}}}{{\rm{f}}}$ = 0

This is required expression

Pure and impure spectrum:

A collection of dispersed light giving its wavelength composition is called a spectrum. If a divergent beam of white light is incident on a prism, then each ray in the beam forms a spectrum. The red of one may coincide with blue of another and the violet of a third and so on. That spectrum is called impure spectrum. The spectrum in which each color distinctly separated from the other is called pure spectrum. Examples for pure spectrum are Hydrogen spectrum and Sodium spectrum. Examples for impure spectrum are Rainbow and Solar spectrum.

When white light having all the wavelengths is passed through an absorbing material, the material may absorb certain wavelengths selectively (to get excited). These wavelengths will disappear when the transmitted light is dispersed. Bark lines or bands at the missing wavelengths one n otherwise bright continuous colored background appear. Such a spectrum is called absorption spectrum. It is of two types: line absorption spectrum and band absorption spectrum. When sunlight’s is superseding, certain sharply defined dark lines are seen. These lines are called Fraunhofer lines.

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